Question

If $\overline{\alpha }$, $\overline{\beta }$ and $\overline{\gamma }$ be vertices of a $△$ whose circumcenter is at the origin, then orthocenter is given by

• A. $\frac{\overline{\alpha }+\overline{\beta }+\overline{\gamma }}{4}$
• B. $\frac{\overline{\alpha }+\overline{\beta }+\overline{\gamma }}{2}$
• C. $\overline{\alpha }+\overline{\beta }+\overline{\gamma }$
• D. $\frac{\overline{\alpha }+\overline{\beta }+\overline{\gamma }}{3}$

Answer 1

The correct option is C $\overline{\alpha }+\overline{\beta }+\overline{\gamma }$

We have $\overline{\alpha }$, $\overline{\beta }$, $\overline{\gamma }$ vertices of a triangle.
Centroid of the $△$ is $\frac{\overline{\alpha }+\overline{\beta }+\overline{\gamma }}{3}$
Now centroid divides the line joining orthocenter and circumcenter in ratio $2:1$
$\therefore$ orthocenter of the $△$ is $\overline{\alpha }+\overline{\beta }+\overline{\gamma }$